Can you explain why you think it fails? The following sample (slightly modified from yours to dump the $s parameter to function):

$str = "abc'\\\\''def";
function f($s) { var_dump($s); return "x"; }
print preg_replace("/c(.*)d/e", "f('\\1')", $str);

outputs:

string(5) "'\\''"
abxef

As it should.

Oddly, I get the following error:

  Warning: Unexpected character in input: '\' (ASCII=92) state=1 in /home/groups/t/ta/tavi/htdocs/playground/test.php(4) : regexp code on line 1

  Parse error: parse error in /home/groups/t/ta/tavi/htdocs/playground/test.php(4) : regexp code on line 1

  Fatal error: Failed evaluating code: f('\'\\\\'\'') in /home/groups/t/ta/tavi/htdocs/playground/test.php on line 4

PHP compile options may be seen at:

  http://tavi.sourceforge.net/playground/phpinfo.php

Ok, I made a little mistake, use "f('\\\\1')" rather than "f('\1')".

Oops, \\1 instead of \1.

H'm...  I'm confused. :-)  To complicate things further, the e-mails display half as many backslashes as the web page. :-)  Since I'm not sure what you mean, I'll exhaustively list the behaviors:

The following:

  $str = "abc'\\\\''def";
  function f($s) { var_dump($s); return "x"; }
  print preg_replace("/c(.*)d/e", "f('\\1')", $str);

gives the error messages I posted earlier.  (That's two backslashes before the 1.)

The following:

  $str = "abc'\\\\''def";
  function f($s) { var_dump($s); return "x"; }
  print preg_replace("/c(.*)d/e", "f('\\\\1')", $str);

gives the result:

  string(2) "\1" abxef

which isn't what I want.  (That's four backslashes before the 1.)

Prefixing the "1" with one or three backslashes in each case emits an ASCII x'01'.

None of which are what I want to occur...

Appears to have been a passing bug in a particular compile.  Cannot repeat.  Closing.